tìm x thuộc Z
\(\left(x+\frac{1}{4}-\frac{1}{3}\right):\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
tìm\(x\in z\)biết :
a.\(\frac{13}{15}-\left(\frac{13}{21}+x\right)\times\frac{7}{12}=\frac{7}{10}\)
b.\(\left(x+\frac{1}{4}-\frac{1}{3}\right)\div\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
Tìm x thuộc Z, biết:
\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)
\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)=\frac{4}{3}.\frac{-1}{3}=\frac{-4}{9}\)
k nha
\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{6}-\frac{3}{6}\right)\)
\(=\frac{4}{3}.\frac{-2}{6}\)
\(=\frac{4}{3}.\frac{-1}{3}\)
\(=\frac{-4}{9}\)
Tìm x:
\(\left(x+\frac{1}{4}-\frac{1}{3}\right)\div\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\left(x+\frac{1}{4}-\frac{1}{3}\right)\div\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\Leftrightarrow\left(x+\frac{1}{4}-\frac{1}{6}\right)\div\frac{23}{12}=\frac{7}{46}\)
\(\Leftrightarrow x+\frac{1}{4}-\frac{1}{6}=\frac{7}{46}\times\frac{23}{12}\)
\(\Leftrightarrow x+\frac{1}{4}-\frac{1}{6}=\frac{7}{24}\)
\(\Leftrightarrow x+\frac{1}{4}=\frac{7}{24}+\frac{1}{6}\)
\(\Leftrightarrow x+\frac{1}{4}=\frac{11}{24}\)
\(\Leftrightarrow x=\frac{11}{24}-\frac{1}{4}\)
\(\Leftrightarrow x=\frac{5}{24}\)
Học tốt !
bài 1: cho x, y thuộc Q. cmr:
|x + y| =< |x| + |y|
bài 2: tính:
\(A=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
bài 3: cho a + b + c = a^2 + b^2 + c^2 = 1 và x : y : z = a : b : c.
cmr: (x + y + z)^2 = x^2 + y^2 + z^2
1
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Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
Bài 3:
Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)
Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a2 + b2 + c2 = 1)
Vậy: (x + y + z)2 = x2 + y2 + z2
tìm x thuộc Z biết :
\(\frac{2}{3}-\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}\cdot\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(\left(x+\frac{1}{4}-\frac{1}{3}\right):\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\left(x+\frac{1}{4}-\frac{1}{3}\right):\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\left(x+\frac{1}{4}-\frac{1}{3}\right):\) \(\frac{23}{12}\) \(=\frac{7}{46}\)
\(x+\frac{1}{4}-\frac{1}{3}\) \(=\frac{7}{45}\times\frac{23}{12}\)
\(x+\frac{1}{4}-\frac{1}{3}\) \(=\frac{161}{540}\)
\(x+\frac{1}{4}\) \(=\frac{161}{450}+\frac{1}{3}\)
\(x+\frac{1}{4}\) \(=\frac{311}{450}\)
\(x\) \(=\frac{311}{450}-\frac{1}{4}\)
\(x\) \(=\frac{397}{900}\)
\(\left(x+\frac{1}{4}-\frac{1}{3}\right):\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\Leftrightarrow\left(x-\frac{1}{12}\right):\frac{23}{12}=\frac{7}{46}\)
\(\Leftrightarrow x-\frac{1}{12}=\frac{7}{24}\)
\(\Leftrightarrow x=\frac{3}{8}\)
Tìm x thuộc Z
a ) \(3\frac{1}{3}:2\frac{2}{5}-1< x< 7\frac{2}{3}.\frac{3}{7}+\frac{5}{7}\)
b) \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)
\(3\frac{1}{3}\div2\frac{2}{5}-1< x< 7\frac{2}{3}\cdot\frac{3}{7}+\frac{5}{7}\)
\(\frac{25}{18}-1< x< \frac{23}{7}+\frac{5}{7}\)
\(\frac{7}{18}< x< \frac{28}{7}\)
\(\frac{49}{126}< x< \frac{504}{126}\)
\(\Rightarrow x=\left(\frac{50}{126};\frac{51}{126};\frac{52}{126};......;\frac{503}{126}\right)\)
tìm x,y biết:
a)\(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)
b)\(x=6:\frac{1}{3}-0,8:\frac{1,5}{\frac{3}{2}.0,4.\frac{50}{1:\frac{1}{2}}}+\frac{1}{4}+\frac{1+\frac{1}{2}.\frac{1}{0,25}}{6-\frac{46}{1+2,2.10}}\)
c)\(y=77^{-1}.7^4.11^2.\left(1.11\right)^4.\left(7^2\right)^{-8}.\left(7^{-8}\right)^{-3}.\frac{1}{\left(-11\right)^{-3}}\)
a) Vì (2x - 5)2000 và (3y + 4)2002 đều có số mũ là chẵn => (2x - 5)2000 \(\ge\) 0; (3y + 4)2002 \(\ge\) 0
Mà tổng trên lại \(\le\) 0
=> (2x - 5)2000 = (3y + 4)2002 = 0
=> 2x - 5 = 3y + 4 = 0
=> x = 2,5; y = \(\frac{-4}{3}\)
b) x = 18 - 0,8 : \(\frac{1,5}{\frac{3}{2}.\frac{4}{10}.\frac{50}{2}}\)+ \(\frac{1}{4}\)+ \(\frac{1+0,5.4}{6-\frac{46}{23}}\)
= 18 - \(\frac{8}{10}:\frac{1,5}{15}+\frac{1}{4}+\frac{3}{4}\)
= \(18-8+1=11\)
a) x = 2,5; y = -4/3
Câu b với c nhìn chóng mặt quá, không dám đụng vào
khó quá à , ai giải được thì nhắn tin cho nhắn nha .
Bài 1:tính giá trị biểu thức sau:
\(B=\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{1326}\right)\)
Bài 2 tìm x biết
\(a,1\frac{1}{30}:\left(24\frac{1}{6}-24\frac{1}{5}\right)-\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(-1\frac{1}{15}\right):\left(8\frac{1}{5}-8\frac{1}{3}\right)\)
\(b,\left|3x-\frac{2}{5}\right|=\frac{1}{35}+\left|\frac{-3}{7}\right|\)
\(c,-3\frac{1}{2}-\left(2x+\frac{1}{2}\right):\left(-\frac{3}{7}\right)=0\)
\(d,\left(x-4\right).\left(x+2\right)\le0\)với x thuộc z
giúp mk với